Question

solve the equation 2cos^2θ-3cosθ+1=0, 0≤θ<0
I believe it has to be in radians. If you could rather help me to the answer, that would be amazing!

Answer 1

\[2\cos ^{2}\theta-3\cos \theta+1=0\]

Answer 2

\[0\le \theta <0\]

Answer 3

can you provide more info like answer choices and what are we doing with this problem explain a little

Answer 4

There are no answer choices

Answer 5

that is alright, so what is the problem asking for us to do?

Answer 6

just to solve for theta

Answer 7

let me ask you how would you solve for x? what steps must taken? I know the answer I want to see if you know I am trying to help here

Answer 8

would it be simplified to \[\cos \theta(\cos \theta+1)=2/3\]?

Answer 9

yes now what?

Answer 10

how do you get theda by itself

Answer 11

This is where I become lost.

Answer 12

wait, would you use arccos to 2/3?

Answer 13

\[\cos \theta+1=\cos^{-1} (2/3)\]

Answer 14

keep going

Answer 15

cos th+1 and sub -1

Answer 16

than divide by cos so you have theda =

Answer 17

so \[\theta=\cos^{-1} (2/3)-1/\]

Answer 18

\[\theta=\cos^{-1} (2/3)-1/\cos\]

Answer 19

yes now it is not done what can be to 2/3-1?

Answer 20

-1/3?

Answer 21

yes so what is the final answer now

Answer 22

Does cos and cos^-1 cancel out leaving only -1/3?

Answer 23

no

Answer 24

Dang.. I do not know what to do after this.

Answer 25

you are done

Answer 26

\[\theta=\cos^{-1} (-1/3)/\cos\]
Nothing need to be placed next to cos?

Answer 27

no

Answer 28

Ok. Thank you :)

Answer 29

did you get what it was asking for my itself? if yes than you are finished

Answer 30

Yes :)